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Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
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c++ code:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> ret;
if(NULL == root) return ret;
queue<TreeNode *> q[2];
stack<vector<int>> s;
int cur=0;
q[cur].push(root);
while(!q[cur].empty()) {
vector<int> tmp;
while(!q[cur].empty()) {
TreeNode *p = q[cur].front();
tmp.push_back(p->val);
if(p->left) q[cur^1].push(p->left);
if(p->right) q[cur^1].push(p->right);
q[cur].pop();
}
cur^=1;
ret.push_back(tmp);
}
return ret;
}
};LeetCode:Binary Tree Level Order Traversal
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原文地址：http://blog.csdn.net/itismelzp/article/details/51524351
